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z^4 + 5z^2 + 4 = 0

z^2 = x

x^2 + 5x + 4 = 0

(x + 4)(x + 1) = 0

x = -4 and -1

z^2 = -4

z^2 = -1

z = +-2i

z + +-i

Its not right; someone else please help!

I guess this should look like:  \(\frac{2x-2}{x^2-x-12}\)

If so, it can be written as \(\frac{2x-2}{(x-4)(x+3)}\)

Express as partial fractions:  \(\frac{2x-2}{(x-4)(x+3)} = \frac{A}{x-4}+\frac{B}{x+3}\)

from which we find    

\(A + B = 2\\\\3A-4B=-2\)

i'll leave you to find A and B.

First, you simplify the sides, pretty straightforward you get x^2 - 3x + 4. Keep in mind that the sum of the roots = -b/a and the product of the roots = c/a. I will put r_1 as a, and r_2 as b. so a + b = -3, and ab = 4. Solve the systems and sub in for a^3 * b + a* b^3

Easy...

We can simplify this into:

a = soccer boots

b = T-shirt

c = cap

a+b = 175

b+c = 64

a = 4c

So we can substitute into:

4c + b = 175

c + b = 64

3c = 111

c = 37

b = 27

a = 148

So the soccer boots cost 148-27 = 121 more than the t-shirt.

(Please tell me if I did something wrong and the answer is incorrect) 

Otherwise there would either be zero solutions or infinite.

Sorry, I am unable to solve your problem. For two variable equations to be solved, there must be two equations.

Since the average is small, we can safely assume that it is easy to find using trial and error.

Row 0's average is 1.

Row 1's average is 1.

Row 2's average is \(\dfrac43\)

Row 3's average is \(\dfrac84\) which is simply 2.

We are not done yet, though.

Row three consists of \(1,3,3,1 \),

so thus \(\boxed{n=3}\)

Just saying maybe you can show me the grid or describe it or something.

Easy...

We can simplify this as 53 - x = 29

We can then take the x to the other side

53 = 29 + x

Then we take the 29 to the other side.

53 - 29 = x

And we get x = 24.

(Please tell me if this answer is wrong) 

Easy...

To get +4x and be the square of some binomial, we can just square (x+2)

Therefore it would be (x+2)^2 

This simplifies to x^2 + 4x + 4

So the answer would be 4.

(Please tell me if I am wrong) 

Here's a diagram,

See the same qustion here:  https://web2.0calc.com/questions/probability-problem_17